Problem 25 - Algebra and Trigonometry

Question

\(21-34\) . Perform the multiplication or division and simplify. $$ \frac{t-3}{t^{2}+9} \cdot \frac{t+3}{t^{2}-9} $$

Step-by-Step Solution

Answer

The simplified expression is \( \frac{1}{t^2+9} \).

Step 1: Write down the expression

We are given the expression \( \frac{t-3}{t^2+9} \cdot \frac{t+3}{t^2-9} \). Our goal is to simplify it.

Step 2: Identify the products

Since we are multiplying two fractions, the result is another fraction where the numerator is the product of the numerators and the denominator is the product of the denominators: \[ \frac{(t-3)(t+3)}{(t^2+9)(t^2-9)} \]

Step 3: Factor the Denominator

The denominator \( t^2-9 \) is a difference of squares. It can be factored as: \[ t^2-9 = (t-3)(t+3) \] So the expression becomes: \[ \frac{(t-3)(t+3)}{(t^2+9)(t-3)(t+3)} \]

Step 4: Cancel common factors

You can cancel \((t-3)(t+3)\) from both the numerator and the denominator, as these are common factors: \[ \frac{1}{t^2+9} \]

Step 5: Final Simplified Expression

After canceling the common terms in the numerator and denominator, the simplified expression is: \[ \frac{1}{t^2+9} \]

Key Concepts

Understanding Multiplication of Fractions

When multiplying fractions, the process is straightforward. You multiply the numerators (the top parts of the fractions) with each other and the denominators (the bottom parts) with each other. Mathematically, for two fractions \( \frac{a}{b} \) and \( \frac{c}{d} \), their product is \( \frac{a \cdot c}{b \cdot d} \). In the exercise provided, we had two fractions, \( \frac{t-3}{t^2+9} \) and \( \frac{t+3}{t^2-9} \), which needed to be multiplied. We performed this by taking the numerator of each fraction and multiplying them together, and doing the same for the denominators:

\[ \frac{(t-3) \cdot (t+3)}{(t^2+9) \cdot (t^2-9)} \]

This step set the stage for further simplification by allowing us to incorporate other algebraic techniques, such as factoring.

Exploring the Difference of Squares

The difference of squares is a specific algebraic pattern which follows the formula \( a^2 - b^2 = (a+b)(a-b) \). Recognizing this pattern makes it easier to factor certain expressions efficiently. In the original exercise, the expression \( t^2 - 9 \) is a perfect example of a difference of squares. Here, \( t^2 \) is the square of \( t \), and \( 9 \) is the square of \( 3 \). Therefore, it can be factored as:
\( (t-3)(t+3) \)
This crucial step allowed us to simplify the original problem further by substituting the factored form into our equation. Such patterns are very useful shortcuts and can dramatically simplify the work involved in algebraic manipulation.

Factoring and Simplification

Factoring is the process of breaking down an expression into a product of simpler expressions, or factors. This step is vital in simplifying complex algebraic fractions. In the exercise, once the product of the fractions was formed, we noticed that the denominator included \( t^2-9 \), which we could factor into \( (t-3)(t+3) \).

Factorization is immensely helpful because, once identified, shared factors in the numerator and denominator, like \( (t-3)(t+3) \), can be "cancelled out". By cancelling these out, we significantly simplify the expression to \( \frac{1}{t^2+9} \).

• This step reduces inappropriate complexity in handling equations.
• It provides simplified expressions for easier interpretation and use.

Remember that factoring can only be done when terms are multiplied, and terms with addition or subtraction have to remain intact.